struct node *reverse(struct node *head, int k )
{
struct node *prev = NULL;
struct node *current = head;
struct node *next = NULL;
int count = 0;
while(current && count < k) {
next = current->next;
current->next = prev;
prev = current;
current = next;
count++;
}
/*** reverse remaining nodes recursively ***/
if (next != NULL) {
head->next = reverse(next, k);
}
return prev;
}
eg
head = reverse(head, 2);
head = reverse(head, 3);
....
-
Azhar.
On Thu, Jun 2, 2011 at 2:09 PM, wujin chen <wujinchen...@gmail.com> wrote:
> how about this one?
>
> Node* reverseBy2(Node* head){
> Node* p1 = head;
> if(p1 == NULL)
> return NULL;
> Node* p2 = p1->next;
> if(p2 == NULL)
> return head;
> Node* nextHead = p2->next;
> p2->next = p1;
> p1->next = reverseBy2(nextHead);
> return p2;
> }
> [?]
>
> 2011/6/1 Shivaji Varma <shivaji...@gmail.com>
>
>> Hi,
>>
>> Please take a look at this link.
>>
>> http://mycsinterviewsexperiences.blogspot.com/
>>
>>
>> --
>> Shivaji
>>
>>
>> On Wed, Jun 1, 2011 at 6:26 AM, Anand <anandut2...@gmail.com> wrote:
>>
>>> Given a linked list of the form, 1->2->3->4->5->6, convert it into the
>>> form 2->1->4->3->6->5. Note that the nodes need to be altered and not the
>>> data contained in them
>>>
>>> --
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>>>
>>
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>>
>
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<<1E3.gif>>
